Heron formula?

I did this exactly the same way as @Chan Lye Lee , but for an alternative approach: let the coordinates of $A$ be $(-10,0)$ and $B(10,0)$ . Then point $C(x,y)$ satisfies $\frac{(x+10)^2+y^2}{21^2}=\frac{(x-10)^2+y^2}{29^2}$

This is a circle with centre $\left(\frac{-641}{20},0\right)$ and radius $\frac{609}{20}$ :

The base of $\Delta ABC$ is fixed ( $AB=20$ ; its height is the $y$ -coordinate of $C$ , which is greatest when $C$ is directly above the centre of the circle; so the largest possible area is $\frac12 \times 20 \times \frac{609}{20} = \boxed{\frac{609}{2}}$

Let $BC=21k$ and $AC=29k$ . Let $s=\dfrac{AB+BC+CA}{2}=25k+10$ . Using Heron's formula, the area of traingle $ABC$ is $\sqrt{s\left(s-AB\right)\left(s-BC\right)\left(s-CA\right)} =\sqrt{\left(25k+10\right)\left(25k-10\right)\left(10+4k\right)\left(10-4k\right)} = 100\sqrt{\left(k^2-\frac{4}{25}\right)\left(\dfrac{25}{4}-k^2\right)}$

By AM-GM inequality, $\sqrt{\left(k^2-\dfrac{4}{25}\right)\left(\dfrac{25}{4}-k^2\right)} \le \dfrac{\left(k^2-\frac{4}{25}\right) + \left(\dfrac{25}{4}-k^2\right)}{2} = \frac{609}{200}$ , which the equality holds if and only if $k^2-\dfrac{4}{25}=\dfrac{25}{4}-k^2$ , that is $k=\sqrt{3.205}$ (achievable).

So, the largest area is $100\times \frac{609}{200} = 304.5$

It is known that, on a plane, the locus of points $C$ such that have a given ratio of distances from two fixed points $A$ and $B$ is is a circle (called a circle of Apollonius ). If $P$ is a point on segment $AB$ such that $\dfrac{\left| \overline{PB} \right|}{\left| \overline{PA} \right|}=r$ and $Q$ is its harmonic conjugate with respect to $B$ and $A$ (i.e. $Q$ is on line $AB$ , outside segment $AB$ such that it has the same ratio of distances from $B$ and $A$ ), then $PQ$ is a diameter of this circle.
In our case, this ratio is $r=\dfrac{21}{29}$ . For simplicity, we ‘ll use the label of any segment to denote both the segment and its length. For instance, $AB=20$ . Let’s calculate the length of this diameter using properties of proportions.

Referring to the figure, $\dfrac{PB}{PA}=\dfrac{21}{29}\Rightarrow \dfrac{PB}{PA+PB}=\dfrac{21}{29+21}\Rightarrow \dfrac{PB}{AB}=\dfrac{21}{50}\Rightarrow PB=\dfrac{21}{50}\times 20\Rightarrow PB=\dfrac{42}{5}$ $\dfrac{QB}{QA}=\dfrac{21}{29}\Rightarrow \dfrac{QB}{QA-QB}=\dfrac{21}{29-21}\Rightarrow \dfrac{QB}{AB}=\dfrac{21}{8}\Rightarrow QB=\dfrac{21}{8}\times 20\Rightarrow QB=\dfrac{105}{2}$ Hence, $PQ=PB+BQ=\dfrac{42}{5}+\dfrac{105}{2}=\dfrac{609}{10}$ Since side $AB$ of $\triangle ABC$ has a fixed length, its area gets maximised, when the corresponding altitude is the longest possible ${{h}_{\max }}$ . Obviously, this altitude is the radius of the Apollonian circle mentioned above, i.e. ${{h}_{\max }}=\dfrac{PQ}{2}=\dfrac{609}{20}$ . Concluding, the largest area that $\triangle ABC$ can have is ${{\left[ ABC \right]}_{\max }}=\dfrac{1}{2}AB\cdot {{h}_{\max }}=\dfrac{1}{2}\times 20\times \dfrac{609}{20}=\boxed{304.5}$